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At a constrained critical point, the directional derivative of $f$ vanishes along every admissible tangent velocity:For a regular zero locus, $T_{a}M=\ker D{g}_a$. A linear functional vanishing on this kernel lies in the row space of $D{g}_a$. Equivalently there are Lagrange multipliers ${\lambda }_1,\dots ,{\lambda }_k$ such thatGeometrically, $\nabla f$ is orthogonal to the tangent space, so it lies in its orthogonal complement, spanned by the independent constraint gradients. Thus the level set of $f$ has no first-order change in a tangent direction. The equations provide candidates for constrained extrema; they must still be classified and compared, and boundary or singular cases must be treated separately. The condition is necessary for a regular constrained extremum, not a guarantee of a maximum.

Since $A$ is invertible, the quadratic form gradient vanishes exactly at the origin. At every nonzero point of $C_{\alpha }$, the scalar derivative is surjective, so the regular level set theorem gives a smooth submanifold of dimension $n-1$.

If $\alpha \ne 0$, the origin is absent. Therefore the largest relatively open smooth subset is all of $C_{\alpha }$.

If $\alpha =0$, the nontriviality assumption forces the form to be an indefinite quadratic form: an invertible definite form would have zero set consisting only of the origin. Every nonzero point is regular. To show that the origin really fails to be a manifold point, use Sylvester's law of inertia to reduce to $|u{|}^2-|w{|}^2=0$. With positive and negative coordinate vectors $e_i,f_j$, the null vectors $e_i+f_j$ and $e_i-f_j$ span ${\mathbb{R}}^n$. Each line in one of these directions lies in the cone.

If a relatively open manifold neighborhood of the origin existed, its tangent space would contain all these line directions, so would have dimension $n$. Such an embedded manifold would be open in ${\mathbb{R}}^n$, impossible for the zero set of a nonzero quadratic form. ConsequentlyThis gives the smooth locus of a nondegenerate real quadratic level set, including maximality at the singular vertex.

The curve-based tangent space at $x$ consists of velocities ${\gamma }^{\prime }(0)$ of smooth curves $\gamma :(-\epsilon ,\epsilon )\to M$ satisfying $\gamma (0)=x$. For an arbitrary subset these velocities need not form a vector space; for a smooth submanifold they do, and this is its usual tangent space. This definition suffices for the direct computation at the identity.

At first order write $\alpha =1+iz$ and $\beta =x+iy$ with real $x,y,z$. The constraint forces the real part of the change in $\alpha$ to vanish. ThenThus the three supplied generators span the tangent space over the reals. Multiplying their matrices givesFor example, $m_1={\sigma }_x/2$, $m_2={\sigma }_y/2$, $m_3=i{\sigma }_z/2$ makes the noncompact signs explicit.

Assume $f$ and $g$ are continuously differentiable near the point. At a regular feasible point, meaning $g(x)=0$ and $\nabla g(x)\ne 0$, a differentiable constrained local extremum satisfies the Lagrange multiplier equationsThey arise by differentiating the optimization Lagrangian $L(x,\lambda )=f(x)-\lambda g(x)$ with respect to all coordinates and the multiplier. Geometrically, the objective has zero derivative along the tangent space of the constraint surface, so its gradient must lie in the normal direction.

Solve these equations for candidates, examine feasible points where the regularity condition fails separately, and compare objective values or use a second-order test restricted to feasible directions. The equations locate stationary points; they do not by themselves establish which are minima or whether a global minimum is attained.

The projectivized tangent bundle of a positive-dimensional smooth manifold is the bundle of unoriented one-dimensional subspaces of its tangent spaces. A local tangent trivialization gives $P(TM){|}_U\cong U\times {\mathbb{RP}}^{n-1}$. If a base-chart transition is $F$, the line coordinates transform by $[v]\mapsto [D{F}_{x}v]$. In any projective chart, this is a ratio of smooth linear expressions with nonzero denominator. It therefore defines a smooth manifold of dimension $2n-1$.

For the two-sphere, a point consists of $x\in S^2$ and a line $L\subset x^{\perp }$. Choosing a unit vector $v\in L$ gives an oriented orthonormal basis $(x,v,x\times v)$. A unique rotation takes one chosen frame to another, so $S{O}_3$ acts transitively. Its stabilizer consists of the identity and the half-turn about $x$, which reverses $v$ while preserving its unoriented line. Thus $P(T{S}^2)\cong S{O}_3/C_2$.

A pseudo-Riemannian manifold is a smooth manifold with a smooth metric tensor that is a nondegenerate bilinear form on each tangent space. Positivity is not required. A Riemannian manifold is the positive-definite case, while a Lorentzian manifold has exactly one temporal direction in its metric signature.

Riemannian geometry studies smooth manifolds equipped with a Riemannian metric: an inner product at each tangent space, varying smoothly with position. This local measuring rule determines angles, arc length, Riemannian distance and geodesics. An inhabitant of a surface can discover its geometry by measuring within it, without knowing how it sits in a larger Euclidean space.

A circular cylinder is a particularly useful example. Its surface bends in space, yet unrolling it preserves the induced metric: a small patch has the geometry of a flat sheet. By contrast, a round sphere has positive Gaussian curvature. The distinction between intrinsic flatness of a circular cylinder and extrinsic curvature separates the geometry an inhabitant measures from the way an outside observer sees the surface bend.

Geodesics supply the analogue of straight lines. On a round sphere they follow great circles; on a circular cylinder they become straight lines when lifted to an unrolled sheet. Their arc length lets us compare candidate routes, as in great-circle distance and Riemannian distance on a circular cylinder. The same measuring rule also defines differential operators such as the Laplace-Beltrami operator, connecting geometry with diffusion and spectral geometry.

Bernhard Riemann's 1854 inaugural lecture proposed studying spaces of arbitrary dimension through their infinitesimal measuring rules, including a positive quadratic expression for the squared line element. This moved the subject beyond a particular surrounding space: the measuring rule itself became part of the geometry. ^[ref].

A Riemannian manifold is a pair $(M,g)$ consisting of a smooth manifold $M$ and a Riemannian metric $g$. For each point $p$, the positive-definite inner product $g_p$ on its tangent space tells us the size and angle of possible instantaneous movements. The smooth manifold describes which local coordinates are possible; the Riemannian metric adds a choice of measurement.

For example, the plane can carry $d{s}^2=d{x}^2+d{y}^2$ or $d{s}^2=4d{x}^2+d{y}^2$. In the second Riemannian metric, a horizontal displacement costs twice as much length as the same coordinate displacement in the first. The underlying smooth manifold is unchanged. The coordinate transformation $X=2x,Y=y$ identifies the second geometry with an ordinary Euclidean plane, illustrating why coordinate appearance alone does not determine curvature.

A round sphere with its induced metric is another Riemannian manifold. Its tangent spaces each have an ordinary Euclidean inner product, although its global geometry is different from a plane's. This is the useful tension: measurements are Euclidean to first order at a point, while how those measurements fit together produces curvature and global restrictions on geodesics.

A Riemannian metric is a smoothly varying positive-definite inner product $g_p$ on each tangent space $T_{p}M$ of a smooth manifold. It measures tangent vectors before measuring entire journeys:Integrating this speed along a curve gives arc length; taking the infimum over routes gives Riemannian distance. A smooth manifold together with this measuring rule is a Riemannian manifold.

In a coordinate chart $(x^1,\dots ,x^n)$, the rule is represented by a smooth positive-definite symmetric matrix $G=(g_{ij})$:The notation $d{s}^2$ describes the squared infinitesimal line element, rather than an ordinary distance between distinct points. Changing coordinates changes the matrix entries; it preserves the measured norm of the same tangent vector.

For the plane with $d{s}^2=4d{x}^2+d{y}^2$, the vector $(1,0)$ has norm $2$ and $(0,1)$ has norm $1$. The coordinate change $X=2x,Y=y$ turns this into $d{s}^2=d{X}^2+d{Y}^2$: coefficients that look different can describe the same geometry. On a circular cylinder the induced metric $R^{2}d{\theta }^2+d{z}^2$ similarly becomes Euclidean when unrolled. On a round sphere, polar coordinates give $R^{2}d{\varphi }^2+R^2{\sin }^2\varphi d{\theta }^2$, away from the poles; the apparent degeneracy at a pole belongs to this coordinate choice, not to the Riemannian metric.

The point of the construction is that measurements make sense intrinsically, even without a chosen surrounding Euclidean space. It gives a common starting point for geodesics, curvature encoded by the Riemann curvature tensor, and the Laplace-Beltrami operator. ^[ref].

A surjective submersion between Riemannian manifolds is Riemannian when its derivative is a linear isometry from the orthogonal complement of its kernel onto the target tangent space. Its vertical space is the kernel and its horizontal space is the orthogonal complement.

If the zero set of a differentiable function $g$ is a submanifold, the chain rule always places its tangent space inside $\ker D{g}_a$. Equality follows from the full-rank hypotheses of the regular level set theorem, but may fail for a singular defining function even when the zero set itself is smooth. For $g:\mathbb{R}\to \mathbb{R}$, $g(x)=x^2$, the zero set is the single point $\{0\}$. Its tangent space is zero, while $D{g}_0=0$ has kernel $\mathbb{R}$. One should therefore check the constraint qualification before using the kernel formula.

The quadratic form gradient gives $D{q}_x(h)=2x^{T}Ah$. Since $A$ is invertible, this linear map is nonzero exactly when $x\ne 0$. The regular level set theorem therefore makes every nonzero point of $C_{\alpha }$ a smooth point of dimension $n-1$. If $\alpha \ne 0$, the entire level set is smooth.

If $\alpha =0$ and the level set contains a nonzero point, the form is an indefinite quadratic form. Its only singular point is the origin. To prove this rather than merely observe a vanishing derivative, use Sylvester's law of inertia to write $q=|u{|}^2-|v{|}^2$ with both sign spaces nonzero. The null vectors $e_i\pm f_j$ span the whole ambient space. Their straight-line curves through the origin lie in the cone. If a relatively open neighborhood of the origin in the cone were an embedded submanifold, its tangent space would contain all these directions and hence have dimension $n$. It would then be open in the ambient space, impossible for the zero set of this nonzero quadratic form. Thus the smooth locus is $C_0\setminus \{0\}$.

Let $Y$ be a embedded submanifold of a complex manifold. If its tangent spaces are invariant under the almost complex structure induced by a complex atlas, choose ambient holomorphic coordinates making its tangent space at one point ${\mathbb{C}}^k\times 0$. The first $k$ coordinates restrict to a local real diffeomorphism by the inverse function theorem, so $Y$ is locally the graph of a smooth function $F$. Invariance of the tangent graph gives $dF(iv)=idF(v)$, the Cauchy-Riemann equations. Thus $F$ is holomorphic and these graph coordinates give the induced complex submanifold structure. This proves the result directly, without needing the general integrability theorem as an unproved step.

For an embedded surface with unit normal vector $\mathbf{n}$, the surface gradient is ${\nabla }_{s}f=(\mathbf{I}-\mathbf{n}{\mathbf{n}}^T)\nabla f$. It projects the ambient gradient onto the tangent space and is independent of how $f$ is extended normally off the surface.

A symplectic basis organizes a nondegenerate alternating bilinear form into independent pairs. Let $(V,\omega )$ be a finite-dimensional vector space over a field $K$, with $\omega$ a nondegenerate bilinear form satisfying $\omega (v,v)=0$ for every vector $v$. Its dimension is even, say $2n$, and a symplectic basis is an ordered basis $\mathcal{B}=(e_1,\dots ,e_n,f_1,\dots ,f_n)$ such thatHere ${\delta }_{ij}$ is the Kronecker delta. Each $(e_i,f_i)$ is a hyperbolic pair; different pairs have zero mutual pairing. These conventions apply in every characteristic. In characteristic two, $-1=1$, but the condition $\omega (v,v)=0$ is still essential: a symmetric bilinear form need not be an alternating bilinear form. The existence theorem and the characteristic caveat are given in Bilinear Forms by Keith Conrad (Keith Conrad, Bilinear Forms, §1, Theorem 1.6; §5, Theorems 5.1 and 5.4 and Definition 5.3).

The benefit is a common coordinate language for every symplectic vector space of the same finite dimension over the same field. The span of all $e_i$, and separately of all $f_i$, is a Lagrangian subspace; each paired span is a symplectic subspace. In Hamiltonian mechanics, the corresponding pairing of position and momentum coordinates describes phase space and canonical transformations. On a symplectic manifold, a symplectic basis describes one tangent space; Darboux coordinates additionally describe a whole local neighborhood.^[ref]

The tangential component is the orthogonal projection of a vector onto the tangent space. For a plane curve, it is $(w\cdot T)T$; for a hypersurface, it is $w-(w\cdot n)n$. The scalar tangential component of acceleration is $\dot{v}$, which can be positive, zero or negative as the speed increases, remains constant or decreases.

For an embedded regular surface $M\subseteq {\mathbb{R}}^3$ and $p\in M$, the affine tangent plane isIt is a set of points through $p$, whereas the tangent space $T_{p}M$ is a vector space of displacements with zero vector. Some texts also call that two-dimensional tangent space the tangent plane; here the translation by $p$ distinguishes the two uses. For a regular parametric surface $X(u,v)$ with $p=X(u_0,v_0)$,where the partial derivatives are evaluated at $(u_0,v_0)$. Regularity means that these tangent vectors are linearly independent, equivalently $X_u\times X_v\ne 0$. Their cross product is a normal vector. ^[ref]

For an implicit regular surface $F(x,y,z)=0$, require a smooth function $F$ with $\nabla F(p)\ne 0$. ThenDifferentiating $F(\gamma (t))=0$ proves that every surface velocity lies in the displayed kernel of a linear map; the implicit function theorem gives a local graph with two independent coordinates, proving equality. The gradient is the normal vector with respect to the ambient dot product. ^[ref]

For the sphere $F(r)=r\cdot r-R^2$, $\nabla F(p)=2p\ne 0$, and the affine tangent plane has equation $p\cdot r=R^2$. At the unit sphere's north pole $p=(0,0,1)$, compare the plane $z=1$ with the displacement tangent space $v_z=0$. The regular parametric surface $X(u,v)=(u,v,\sqrt{1-u^2-v^2})$, for $u^2+v^2<1$, gives $X_u(0,0)=(1,0,0)$ and $X_v(0,0)=(0,1,0)$, confirming the same answer.

This analytical illustration shows a unit sphere and a vertical cross-section. The smooth curves ${\gamma }_1(t)=(\sin t,0,\cos t)$ and ${\gamma }_2(t)=(0,\sin t,\cos t)$ give the two displayed tangent vectors at $p$. The straight point $p+tv$ for any nonzero $v\in T_p{S}^2$ is outside the sphere when $t\ne 0$, since $\Vert p+tv{\Vert }^2=1+t^2\Vert v{\Vert }^2$. A tangent plane therefore captures first-order motion, not an exact patch of the regular surface.

The same idea makes local approximation useful. For the graph of a function $z=f(x,y)$ with $f$ a smooth function, the affine tangent plane is $z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)$. ^[ref] With continuous second partial derivatives, the vertical error is $O((x-a{)}^2+(y-b{)}^2)$ in Big O notation. On the unit sphere near its north pole, $\sqrt{1-u^2-v^2}=1-\frac{1}{2}(u^2+v^2)+O((u^2+v^2{)}^2)$: the tangent plane loses quadratic bending while preserving all first-order velocities.

A failed regularity test needs interpretation. Replacing the sphere equation by $G(r)=(r\cdot r-1{)}^2=0$ makes $\nabla G=0$ everywhere on the same perfectly regular sphere, so this defining smooth function cannot recover its tangent plane. Similarly, the usual spherical parametric surface loses matrix rank at the poles, while the graph parametrization above remains regular there. A vanishing gradient or degenerate parametrization invalidates that calculation; it alone does not prove that the underlying geometry has no tangent plane.

At a fixed point $p$ of an $n$-dimensional smooth manifold $M$ without boundary, the tangent space $T_{p}M$ is the real vector space of all tangent vectors at $p$. In manifold charts $x=(x^1,\dots ,x^n)$ and $y=(y^1,\dots ,y^n)$, a tangent vector has componentsThe coordinate basis transforms inversely, so the tangent vector is unchanged. This invertible Jacobian matrix also preserves vector addition and scalar multiplication, giving a vector space of dimension $n$ independently of coordinates. ^[ref]

For example, at $(x,y)=(1,0)$ in ${\mathbb{R}}^2$, use the alternative manifold chart $u=x$, $w=y+x^2$. The Jacobian matrix sends $(v_x,v_y)=(2,3)$ to $(v_u,v_w)=(2,7)$. Indeed, the smooth curve $(x(t),y(t))=(1+2t,3t)$ gives $w(t)=1+7t+4t^2$. Conversely, ${\partial }_u{|}_p={\partial }_x{|}_p-2{\partial }_y{|}_p$ and ${\partial }_w{|}_p={\partial }_y{|}_p$, so $2{\partial }_u{|}_p+7{\partial }_w{|}_p=2{\partial }_x{|}_p+3{\partial }_y{|}_p$. The numbers change; their action on every smooth function does not.

For an embedded submanifold $M\subseteq {\mathbb{R}}^N$, the tangent space can be identified with the vector subspaceA proof outline explains why every coordinate vector is possible: in a manifold chart $x$, lift the short straight path $x(p)+ta$ using $x^{-1}$. Its tangent vector has components $a$. For a regular local parametrization $X:U\subseteq {\mathbb{R}}^n\to M$, the same chain rule gives $T_{p}M=\mathrm{im}D{X}_q$, where $X(q)=p$. This is the tangent space from a local parametrization.

For the sphere $S_R^2=\{p\in {\mathbb{R}}^3:p\cdot p=R^2\}$, differentiating that equation gives $p\cdot v=0$. Conversely, for any nonzero $v$ with $p\cdot v=0$, the smooth curvelies on the sphere and has ${\gamma }_v^{\prime }(0)=v$; use a constant smooth curve for $v=0$. Thus $T_p{S}_R^2=\{v:p\cdot v=0\}$, not merely a subset of that plane. At $p=(0,0,R)$ this vector subspace is $v_z=0$, while the affine tangent plane drawn through $p$ is $z=R$.

This also explains a mechanical constraint without requiring a force calculation. In the ideal mathematical model of a particle confined to a stationary sphere, its allowed velocities at $p$ are exactly $T_p{S}_R^2$: the radial component must vanish, but either independent tangential direction is available. For a general stationary holonomic constraint $F(q)=0$ with independent equations, the regular level set theorem gives $T_{q}M=\ker D{F}_q$. These are instantaneous kinematic possibilities; they do not determine the actual subsequent motion. On a general smooth manifold, tangent spaces at different base points cannot be identified canonically; an affine connection supplies parallel transport when such comparisons are needed.

A tangent vector at a point $p$ of a smooth manifold without boundary records an instantaneous velocity through $p$. Its base point is part of the data; measuring its length additionally requires a Riemannian metric. Intrinsically, a smooth curve $\gamma$ with $\gamma (0)=p$ defines its tangent vector by how every local smooth function changes along it:This is a derivation at a point. Two smooth curves through $p$ represent the same tangent vector if these derivatives agree for every such smooth function; equivalently, their coordinate velocity vectors agree in one manifold chart. The chain rule makes that agreement independent of the chosen manifold chart. ^[ref]

For an embedded submanifold of Euclidean space, the intrinsic tangent vector is represented by the ordinary derivative ${\gamma }^{\prime }(0)$, with its base point $p$ retained. A drawn arrow starting at $p$ represents that tangent vector; its endpoint $p+{\gamma }^{\prime }(0)$ need not lie on the embedded submanifold. Nor is a tangent vector required to be a unit tangent vector: a stationary smooth curve gives the zero vector, which belongs to the tangent space but has no direction to normalize.

Here are explicit calculations. In ${\mathbb{R}}^2$, $\gamma (t)=(t,t^2)$ and $\eta (t)=(t,2t^2)$ have different routes but the same tangent vector $(1,0)$ at $(0,0)$. For $f(x,y)=3x+y$, both give $v(f)=3$. For a circle of radius $R$, $c(t)=(R\cos (\omega t),R\sin (\omega t))$ has tangent vector $(0,R\omega )$ at $(R,0)$: changing $\omega$ changes the speed or reverses travel without changing the circle.

A reparametrization $\beta (s)=\gamma (h(s))$ with $h(0)=0$ gives, by the chain rule, ${\beta }^{\prime }(0)=h^{\prime }(0){\gamma }^{\prime }(0)$. A regular change of curve parameter has $h^{\prime }(0)\ne 0$. The singular choice $h(s)=s^3$ instead givesBoth describe the same parabola, whose tangent line at the origin is still the horizontal axis. The zero velocity diagnoses this singular change of curve parameter, whose inverse is not differentiable at zero; it does not show that the geometric tangent line is absent. The regular curve $\gamma$ supplies its nonzero direction.

The diagram is an analytical illustration of these exact parametric curves, with equal parameter increments marked on each route. It separates geometry from traversal: the tangent line depends on the route, while a velocity vector also depends on how the curve parameter advances.